added problem of stack

Author: muskanmi

Stack/MaxRectangle.java

@@ -0,0 +1,137 @@
+package Stack;
+
+public class MaxRectangle {
+
+    public static int[] nextsmallerelement(int[] arr, int n)
+  {
+ 
+    Stack<Integer> st = new Stack<>();
+ 
+    // For the elements which dont have
+    // next smaller element ans will be -1
+    st.push(-1);
+ 
+    // Store indices in output
+    int[] right  = new int[n];
+ 
+    // Start from last index
+    for (int i = n - 1; i >= 0; i--) {
+ 
+      // If top element is sorted then
+      // no need to do anything, just store
+      // the answer and push the
+      // current element in stack
+      if ((st.peek() != -1)
+          && arr[st.peek()] < arr[i]) {
+        right[i] = st.peek();
+        st.push(i);
+      }
+      else {
+        while ((st.peek() != -1)
+               && arr[st.peek()]
+               >= arr[i]) {
+          st.pop();
+        }
+        right[i] = st.peek();
+        st.push(i);
+      }
+    }
+    return right;
+  }
+
+    public static int[] previousmallerelement(int arr[],int n)
+    {
+      Stack<Integer> st = new Stack<>();
+      st.push(-1);
+   
+      int[] left  = new int[n];
+   
+      // Start from first index - Difference Between Next and Previous Smaller element
+      for (int i = 0; i < n; i++) {
+        if ((st.peek() != -1)
+            && arr[st.peek()] < arr[i]) {
+          left[i] = st.peek();
+          st.push(i);
+        }
+        else {
+          while ((st.peek() != -1)
+                 && arr[st.peek()]
+                 >= arr[i]) {
+            st.pop();
+          }
+          left[i] = st.peek();
+          st.push(i);
+        }
+      }
+      return left;
+    }
+    public static int getMaxArea(int [] arr, int n)
+  {
+    int [] right = new int [n];
+    right = nextsmallerelement(arr, n);
+ 
+    // Find the smallest element than
+    // curr element in right side
+ 
+    int [] left = new int [n];
+    left = previousmallerelement(arr, n);
+ 
+    // Find the smallest element
+    // than curr element in left side
+    int maxarea = Integer.MIN_VALUE;
+ 
+    // Now the left and right array have
+    // index of smallest element in left and
+    // right respectively, thus the difference
+    // of right - left - 1 will give us
+    // breadth and thus
+    // area = height(curr==arr[i]) * breadth;
+    for (int i = 0; i < n; i++) {
+      int height = arr[i];
+      if (right[i] == -1) {
+        right[i] = n;
+      }
+      int breadth = right[i] - left[i] - 1;
+      maxarea = Math.max(maxarea,
+                         height * breadth);
+    }
+    return maxarea;
+  }
+
+    public static int maxRectangleArea(int [][] M, int R, int C)
+    {
+      int area = getMaxArea(M[0], R);
+      int maxarea = area;
+   
+      for (int i = 1; i < R; i++) {
+        for (int j = 0; j < C; j++) {
+          if (M[i][j] != 0) {
+   
+            // Add heights of previous rows
+            // into current
+            M[i][j] = M[i][j]
+              + M[i - 1][j];
+          }
+          else {
+   
+            // If current height is 0 then
+            // don't add previous heights
+            M[i][j] = 0;
+          }
+        }
+        maxarea = Math.max(maxarea,
+                           getMaxArea(M[i], R));
+      }
+      return maxarea;
+    }
+    public static void main(String[] args) {
+        int R = 4, C = 4;
+        int [][]amt = {
+          { 0, 1, 1, 0 },
+          { 1, 1, 1, 1 },
+          { 1, 1, 1, 1 },
+          { 1, 1, 0, 0 },
+        };
+        System.out.println(maxRectangleArea(amt, R, C));
+    }
+}